Problem: $f(x, y, z) = zx^2 - zy^2 + e^y\cos(z)$ Is $f$ harmonic? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Solution: A scalar field $f$ is harmonic if its Laplacian is zero. The Laplacian of a scalar field $f$ is the sum of each of its second partial derivatives. $\Delta f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2}$ [What does that triangle mean?] Let's find the second partial derivatives of $f$ ! $\begin{aligned} f_{xx} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ 2xz \right] \\ \\ &= 2z \\ \\ f_{yy} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ -2yz + e^y\cos(z) \right] \\ \\ &= -2z + e^y\cos(z) \\ \\ f_{zz} &= \dfrac{\partial}{\partial z} \left[ \dfrac{\partial f}{\partial z} \right] \\ \\ &= \dfrac{\partial}{\partial z} \left[ x^2 - y^2 - e^y\sin(z) \right] \\ \\ &= -e^y\cos(z) \end{aligned}$ The Laplacian: $\Delta f = 2z - 2z + e^y\cos(z) - e^y\cos(z) = 0$ Because the Laplacian of $f$ is zero, $f$ is harmonic.